The denominators of these fractions are the same, so the fractions are the same if their numerators are equal.

This means we need to find A and B so that

A(x + 5) + B(3x + 2) = 7x + 9.

There's an easy way and a harder way to do this. The harder way is to multiply out the left-hand side, find that

(A + 3B)x + (5A + 2B) = 7x + 9,

and solve the system of equations

A + 3B = 7

5A + 2B = 9.

We don't want to do that, because it's too much work. Instead, we'll carefully choose values of x so that we're solving for only one of A or B at a time.

If we take x = -5, for example, then we don't have to worry about A.

A(x + 5) + B(3x + 2) = 7x + 9

A((-5) + 5) + B(3(-5) + 2) = 7(-5) + 9

A(0) + B(-13) = -26

B = 2.

Now that we know B = 2, we can use any value of x we like to solve for A. Let's take x = 0.

A(x + 5) + B(3x + 2) = 7x + 9

A((0) + 5) + (2)(3(0) + 2) = 7(0) + 9

5A + 4 = 9

5A = 5

A = 1.

Now we know that A = 1 and B = 2, so we've successfully decomposed the original fraction:

As long as the denominator of the original function factors into distinct linear factors, there will be one partial fraction for each factor of the denominator. For the time beingwe'll be nice and give you the original function with the denominator already factored.

Example 2

Decompose

into partial fractions.

There will be one partial fraction for each factor of the denominator, so we want to find A and B such that

Adding the partial fractions, we get

These
fractions have the same denominator, so they will be equal as long as
their numerators are equal. This means we need to find A and B such that

7x + 5 = A(x – 1) + B(x + 2).

We want to find A and B the easy way. Choose x = 1 so that we can find B without worrying about A:

7x + 5 = A(x – 1) + B(x + 2)

7(1) + 5 = A((1)-1) + B(1 + 2)

12 = 3B

B = 4.

Now that we know B, we can take any value of x we like and find A. Take x = 0: